Problem 88 - Chemistry and Chemical Reactivity

Question

A A tree is exactly \(10 \mathrm{m}\) tall. (a) What must be the total molarity of the solutes if sap rises to the top of the tree by osmotic pressure at \(20^{\circ} \mathrm{C}^{2}\) Assume the groundwater outside the tree is pure water and that the density of the sap is \(1.0 \mathrm{g} / \mathrm{mL} .\left(1 \mathrm{mm} \mathrm{Hg}=13.6 \mathrm{mm} \mathrm{H}_{2} \mathrm{O} .\right)\) (b) If the only solute in the sap is sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) what is its percent by mass?

Step-by-Step Solution

Answer

(a) 0.0402 M, (b) 1.376%

Step 1: Understanding Osmotic Pressure

Osmotic pressure is the pressure required to stop the flow of water through a semipermeable membrane due to osmosis. It can be calculated using the formula: \[ \Pi = iMRT \] where \( \Pi \) is the osmotic pressure, \( i \) is the van 't Hoff factor (which is 1 for non-ionizing solutes like sucrose), \( M \) is the molarity, \( R \) is the ideal gas constant \( (0.0821 \text{ L atm/mol K}) \), and \( T \) is the temperature in Kelvin (293 K for \( 20^{\circ} \text{C} \)).

Step 2: Calculating Osmotic Pressure

The osmotic pressure needs to be enough to raise the sap to the height of the tree. The column of sap has to exert a pressure equal to the column of water it displaces. The pressure required to lift a column of water \( h = 10 \text{ m} \) can be found in millimeters of Hg using the conversion: \[ h_{\text{Hg}} = \frac{h_{\text{H}_2\text{O}} \times 1 \text{ mmHg}}{13.6} \] Given \( h_{\text{H}_2\text{O}} = 10000 \text{ mm} \), \[ h_{\text{Hg}} = \frac{10000}{13.6} \approx 735.29 \text{ mmHg} \] Convert mmHg to atm: \[ 735.29 \times \frac{1 \text{ atm}}{760 \text{ mmHg}} \approx 0.9675 \text{ atm} \]

Step 3: Determining Molarity using Osmotic Pressure

Rearrange the osmotic pressure formula to solve for molarity \( M \): \[ M = \frac{\Pi}{iRT} = \frac{0.9675}{1 \times 0.0821 \times 293} \approx 0.0402 \text{ M} \] Thus, the total molarity of the solutes necessary for the sap to rise to the top of the tree is 0.0402 M.

Step 4: Calculating Mass Percent of Sucrose

First, calculate the molar mass of sucrose \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \): \( 12 \times 12.01 \text{ g/mol} + 22 \times 1.01 \text{ g/mol} + 11 \times 16.00 \text{ g/mol} = 342.34 \text{ g/mol} \) Calculate the mass of sucrose in 1 L of sap: \[ 0.0402 \text{ moles/L} \times 342.34 \text{ g/mol} = 13.76 \text{ g/L} \] Since the sap density is 1.0 g/mL or 1000 g/L, \[ \text{Mass percent} = \left( \frac{13.76 \text{ g}}{1000 \text{ g}} \right) \times 100 \approx 1.376\% \]

Key Concepts

Van 't Hoff Factor

The Van 't Hoff factor, often symbolized as \( i \), plays a crucial role in determining the osmotic pressure of a solution. It accounts for the effect of solute particles in solutions and is defined as the number of particles a solute creates in solution. For non-ionizing solutes, such as sucrose, which do not dissociate in water, \( i = 1 \). This indicates that for every molecule of sucrose dissolved, one solute particle is produced.

In scenarios involving ionic solutes, this factor would differ because ionic substances dissociate in solution. For instance, when sodium chloride (NaCl) dissolves, it separates into two ions, resulting in an \( i \) value of 2. Therefore, understanding the Van 't Hoff factor helps predict how a solute will affect a solution's colligative properties, including osmotic pressure.

• Non-ionizing solute (like sucrose): \( i = 1 \)
• Ionic solute example (like NaCl): \( i = 2 \)

Molarity

Molarity (\( M \)) is a measurement of concentration, denoting the number of moles of solute per liter of solution. It is pivotal in chemical solutions as it provides a way to express the concentration of solutes in a consistent and useful manner. In calculating osmotic pressure using the formula \( \Pi = iMRT \), molarity directly influences how much pressure a solute exerts.

In our tree example, the necessary molarity to lift sap 10 meters high was calculated to be \( 0.0402 \text{ M} \). This shows that to generate sufficient osmotic pressure for this rise, a concentration of 0.0402 moles of sucrose per liter is required. Such calculations consider various conversion factors and constant values, like the ideal gas constant \( R = 0.0821 \text{ L atm/mol K} \).

• Molarity determines the strength of the osmotic effect.
• It is measured in moles per liter (mol/L).

Sap Density

Knowing the density of sap is essential for comprehending how plants transport nutrients and maintain structure. Sap density is given as \( 1.0 \text{ g/mL} \), equivalent to \( 1000 \text{ g/L} \), providing insight into the mass percent of solutes like sucrose in the sap. This information is vital when calculating the osmotic pressure needed for sap to rise within a tree.

For example, with the molarity of sucrose calculated, we estimated the mass of sucrose in one liter of sap to be approximately \( 13.76 \text{ g} \). Given the sap's density, this converts to about 1.376% sucrose by mass. Such calculations require a clear understanding of how density relates to volume and mass, especially in solutions where density can determine solute concentration.

• Sap density influences concentration calculations.
• This density helps parameterize the problem, connecting mass with volume.